Answer:
The Wife is a carrier of the recessive allele with a genotype XX°
The husband is a colorblind individual with Genotype X°Y
Explanation:
- We know that Color blindness is a Sex linked recessive disorder.
Now let us consider the case of the colorblind daughter ,
- She must be homozygous for the recessive allele because a hetrozygous female will not show the phenotype of the x-linked recessive trait.
- Hence , The color blind daughter must have a genotype X°X°.
- This clearly shows that she has received one mutated X-chromosome from her mother and One mutated chromosome from her Father.
- Since, the father has the Mutated gene he must be a Color blind individual and has a X°Y genotype.
- The Mother could either be a Carrier or a colorblind individual.
Now, Let us consider the case of a Normal Boy:
- Sons always receive their X chromosome from mother and genes present on the X-chromosome of males always express themselves.
- Absence of the disease in this son states that he has received a normal X-chromosome from the mother.
On considering both cases stated above we can conclude that the woman is a carrier of the recessive trait while father expresses the disease and has the X-chromosome with the recessive allele.
This is further confirmed by presence of a heterozygous daughter whose genotype is XX°. She has received the normal X from mother and the mutated X from the father.
